the sum of the divisors of $2^5{.3}^7{.5}^3{.7}^2$  is

Any divisor of $2^5.3^7.5^3.7^2$  is of the form $2^a{3}^b{5}^c{7}^d,$  where $0\le b\le 7,0\le c\le 3$  and $0\le d$  $\le 2.$  

Thus, the sum of the divisors of $2^5.3^7.5^3.7^2$  is

            $(1+2+\mathrm{...}+2^5)(1+3+\mathrm{...}+3^7)$ $(1+5+5^2+5^3)(1+7+7^2)$

           $=\left(\frac{2^6-1}{2-1}\right)\left(\frac{3^8-1}{3-1}\right)\left(\frac{5^4-1}{5-1}\right)\left(\frac{7^3-1}{7-1}\right)$

         $=\frac{(2^6-1)(3^8-1)(5^4-1)(7^3-1)}{48}$